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Homework Help: Radius of Convergens

  1. Apr 18, 2006 #1

    I'm told that the the power series:

    [tex]\sum_{n=0} ^ \infty (2n+1) z^n[/tex] has the radius of convergens

    R = 1.


    Using the Definition of convergens for power series:

    [tex]\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)} [/tex]

    [tex]limit _{n \rightarrow \infty} \frac{(2n+1)}{(2n+3)} = 1[/tex]

    Therefore the radius of convergens is R = 1. Right ?

    Second question: The Power series above suposedly diverges on every point on the circle of convergens. How do I show that?

    I know that according to the definition of divergens of the power series:

    [tex]\sum_{n = 0} ^{\infty} a_n z^n [/tex]

    that [tex]a_n \rightarrow \infty [/tex] for [tex]n \rightarrow \infty [/tex]

    Do I the use this fact here to show that a_n diverges ??

    Best Regards
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2
    I'm not following this step. I would use the ratio test for finding the radius of convergence.
  4. Apr 20, 2006 #3
    I think this would be the proper way to evaluate the series:

    [tex]\sum_{n=0} ^ \infty (2n+1) z^n[/tex]

    so using the ratio test,

    [tex]limit _{n \rightarrow \infty} \frac{(2(n+1)+1) z^(n+1)}{(2n+1) z^n} [/tex]

    [tex]limit _{n \rightarrow \infty} \frac{(2n+3)}{(2n+1)}|z| [/tex]

    Therefore, [tex]|z|<1[/tex]

    implies that the interval of convergence is [tex] -1<z<1 [/tex]

    Therefore, the radius of convergence must be 1.
  5. Apr 23, 2006 #4
    Hello and thank Your for Your answer,

    Anyway if I then have show that the series diverges for all point on the circle of convergens.

    Doesn't that mean that

    [tex]\frac{2n+3}{2n+1}|z| \geq 1[/tex], where [tex]n \neq 0[/tex]

    if n = 1

    then [tex]|z| \geq \frac{3}{5}[/tex]

    Am I on the right track here?

    Best Regards

    Last edited: Apr 23, 2006
  6. Apr 23, 2006 #5


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    Science Advisor

    n= 1 is not relevant. What is
    [tex]lim_{n\rightarrow \infty}\frac{2n+3}{2n+1}[/tex]?
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